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Question #47401
The Question
Allen’s Rule states that mammals from colder climates generally have shorter ears and limbs to minimize heat loss. When analyzing the appendicular skeleton and thermoregulatory adaptations of a polar mammal compared to a closely related desert mammal, which of the following statements is INCORRECT?
Options
- A.
The polar mammal will exhibit a lower surface area to volume (SA:V) ratio in its extremities, thereby reducing the rate of convective heat dissipation.
- B.
The distal appendicular bones (e.g., radius, ulna, tibia, fibula) of the polar mammal are relatively shortened, representing an evolutionary trade-off that prioritizes heat retention over maximum stride length.
- C.
To maximize core heat retention, the polar mammal develops a highly expanded appendicular girdle to support elongated phalanges, keeping the vital organs further from the freezing ground.
- D.
The desert mammal utilizes its elongated appendicular skeleton as a thermal window, facilitating a higher SA:V ratio to maximize heat loss in a high-temperature environment.
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Correct Answer
Step-by-Step Solution
Explanation
How the Examiner Thinks
Comparative_Analysis
Sharpen your ability to evaluate multiple scenarios and distinguish between competing potential answers.
Concepts Behind This Question
Appendicular Girdle|Allen’s Rule|Surface-to-Volume Ratio
Why This Question Is Tricky
Common Mistakes
Conceptual Pivot: Students often fall for the 'Common Half-Truth' trap.
While keeping the core away from freezing ground sounds logically advantageous for a polar animal, it ignores the overriding physiological constraint of Allen's Rule.
Students must recognize that any elongation of the distal appendicular skeleton (phalanges) fundamentally violates the principle of minimizing the SA:V ratio for heat retention.
Quick Facts About This Question
Question Information
| Subject | Biology |
| Chapter | Locomotion and Movement - Class 11 |
| Concept | Appendicular Girdle Mechanics |
| Difficulty | 8 |
| Cognitive Load | 5 |
| Target Solve Time | 120 sec |
| Access | Free |